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3

This has been fixed in 10.0.2. The longer time now remain in the status windows. On windows 7, 64 bit SetOptions[$FrontEnd, EvaluationCompletionAction -> "ShowTiming"] Plot[{BesselJ[1, x], BesselJ[2, x]}, {x, 0, 10}, PlotPoints -> 1*^5, Filling -> {1 -> {2}}]


1

Trick that I sometimes find useful is "inheriting" of default option value from another symbol using delayed rule like: optName :> OptionValue[anotherSymbol, optName] With f defined as: ClearAll[f]; Options[f] = {optA -> 1, optB -> 1, optC -> 1}; f[x_, opts : OptionsPattern[]] := OptionValue[{optA, optB, optC}] We can define g in following ...


13

Normally I like to use On and Off for this kind of tracing as it is easy to set up without modifying any symbols. However, it does not immediately work in this case: On[Roots] Solve[x^3 - 2 x + 12 == 0, x]; Off[] This does not produce any trace messages. Something must be using Quiet to suppress them. We can check this hypothesis: On[Quiet] Solve[x^3 ...


15

Let me first answer your second question, since I can only guess about the main question: I also observed that the syntax colouring (version 10, windows 7) suggests that Trace can be used with only two arguments. It's really just the coloring that goes wrong and has nothing to do with functionality. You can see that it is not even related to ...


1

Restrict the points using VectorPoints. Clear[myPoints, myVectors]; myPoints = Take[Select[ RandomVariate[ UniformDistribution[{{-1, 1}, {-1, 1}}], {1000}], .3 < Norm[#] < 1. &], 500]; myVectors = Table[{Cos[\[Theta] = RandomReal[{0, 2 \[Pi]}]], Sin[\[Theta]]}, {Length[myPoints]}]; ListVectorPlot[ Transpose[{myPoints, ...


2

I saw this somewhere else on here, but I cannot remember where. The simplest method is to use a fresh kernel: SetOptions[Plot, Axes -> False]; LaunchKernels[1]; ParallelEvaluate[Options[Plot, Axes]] (* {{Axes -> True}} *) then you dispose of the kernel AbortKernels[] (* {KernelObject[1, "local"]} *)


4

Another option is to use Trace to find occurrences of the relevant rules which appear during evaluation: Trace[ Plot[Sin[t], {t, 0, 2 Pi}], HoldPattern[PlotPoints | MaxRecursion -> _], TraceInternal -> True] // Flatten // Union {MaxRecursion -> 6, MaxRecursion -> Automatic, PlotPoints -> 50, PlotPoints -> Automatic} Or you can set "Verbose" ...


6

$Version "10.0 for Mac OS X x86 (64-bit) (September 10, 2014)" From the documentation at http://reference.wolfram.com/language/tutorial/Options.html the default for PlotPoints is 50. To determine the default for MaxRecursion, let f[x_] = Product[x + n (-1)^n, {n, -4, 5}] E^(-x^2/2); The number of points for the Plot of f[x] with the defaults for ...


3

When all else fail, i.e. Options[p1] or AbsoluteOptions[p1, PlotPoints], you can always grab the Line itself and find how many points it has: p1 = Plot[Sin[x], {x, 0, Pi}, PlotPoints -> Automatic] line = p1[[1, 1, 3, 2]]; Graphics[line] Length[line[[1]]] So, 259 points are used in this case. ListPlot[line[[1]], Mesh -> All, PlotStyle ...


2

You are fooled by the fact that the vertical axis starts from 0.87, while you expect it to start from 0:


1

The central problem is that AspectRatio governs the full figure, including the Ticks, which extend beyond the graph. You can get the proper AspectRatio for a graph that does not contain the Ticks as follows: basePlot = Plot[PDF[BetaDistribution[0.8, 0.8], x], {x, 0, 1}, Filling -> Axis, Ticks -> None]; myAspectRatio = #[[2]]/#[[1]] & ...


5

Stealing Lou's example, you can add a couple of options: CreateDocument[ Manipulate[ Plot[Sin[x (1 + a x)], {x, 0, 6}, ImageSize -> Full], {a, 0, 2}], CellMargins -> 0, ShowCellBracket -> False, WindowSize -> Full] If you want those options to be always valid, you can put SetOptions[EvaluationNotebook[],...] into ...



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